Question

The number $21!$ $=51{,}090{,}942{,}171{,}709{,}440{,}000$ has over $60{,}000$ positive integer divisors. One of them is chosen at random. What is the probability that it is odd?

$\dfrac{1}{21}$

$\dfrac{1}{19}$

$\dfrac{1}{18}$

$\dfrac{1}{2}$

$\dfrac{11}{21}$

Answer: B

Concepts:factor counting prime factorization basic probability

Difficulty rating: 1540

Solution:

The exponent of $2$ in $21!$ is $\begin{aligned} &\left\lfloor\dfrac{21}{2}\right\rfloor+ \left\lfloor\dfrac{21}{4}\right\rfloor\\ &\quad{}+\left\lfloor\dfrac{21}{8}\right\rfloor+ \left\lfloor\dfrac{21}{16}\right\rfloor\\ &=10+5+2+1=18. \end{aligned}$ Thus $21!=2^{18}d$ for some odd integer $d.$

For every divisor $x$ of $d,$ the divisors of $21!$ with odd part $x$ are $x,2x,\ldots,2^{18}x.$ Exactly one of these $19$ divisors is odd, so the probability is $\dfrac1{19}.$

Thus, the correct answer is B .

2017 AMC 10B Problem 20

Answer: B

Difficulty rating: 1540

Solution:

Problem 20 in Other Years